A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a 900.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid sodium hydroxide and add it to the flask. Fill the flask to the mark with distilled water.

28.58 g of NaOH

Explanation:

The question is incomplete. The missing part is:

"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"

To do this, we need to know how much of the base we have to weight to prepare this solution.

First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:

NaOH > Na⁺ + OH⁻

As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.

This can be done with the following expression:

14 = pH + pOH

and pOH = -log[OH⁻]

So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:

pOH = 14 - 13.9 = 0.10

[OH⁻] = 10⁽⁻⁰°¹⁰⁾

[OH⁻] = 0.794 M

Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:

n = M * V

n = 0.794 * 0.9

n = 0.7146 moles

Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:

m = 39.997 * 0.7146

dynamic equilibrium is the answer.